To solve the equation \( 729^{4x-1} = 9^{3x+8} \) using the laws of exponents, we start by rewriting both sides in terms of base \(3\).
We know that:
- \( 729 = 3^6 \) (since \( 9^3 = (3^2)^3 = 3^6 \))
- \( 9 = 3^2 \)
We can rewrite the left side: \[ 729^{4x-1} = (3^6)^{4x-1} = 3^{6(4x-1)} = 3^{24x - 6} \]
Now, we'll rewrite the right side: \[ 9^{3x+8} = (3^2)^{3x+8} = 3^{2(3x+8)} = 3^{6x + 16} \]
Now we have the equation: \[ 3^{24x - 6} = 3^{6x + 16} \]
Since the bases are the same, we can set the exponents equal to each other: \[ 24x - 6 = 6x + 16 \]
Now, we will solve for \(x\):
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Subtract \(6x\) from both sides: \[ 24x - 6x - 6 = 16 \] \[ 18x - 6 = 16 \]
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Add \(6\) to both sides: \[ 18x = 16 + 6 \] \[ 18x = 22 \]
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Divide by \(18\): \[ x = \frac{22}{18} = \frac{11}{9} \]
None of the provided choices exactly match \( \frac{11}{9} \). However, let's confirm whether typographical errors exist regarding \(\frac{19}{6}\) and \(\frac{11}{6}\).
Since we derived \(x = \frac{11}{9}\), and it does not match the available options, please double-check the choices or consider if a revision or reevaluation is necessary.
Final value: \[ x = \frac{11}{9} \]
1 answer